x3 6x2-x-30
In the absence of a proper statement of the equation I can only guess at answers. How about (2x + 3)(3x - 5) which is the factorisation of 6x2 - x - 15 = 0? In which case x = -2/3 or 5/3.
6x2-24=0 6(x2-4)=0 x= {-2,2}
Yes.(2x)2 = 2*2*x*xor2x2 = 2 * x * x
3/(x2+4x-1) = 6/(2x2-3x+5) Cross - multiply in order to eliminate the fractions: 6*(x2+4x-1) = 3*(2x2-3x+5) 6x2+24x-6 = 6x2-9x+15 6x2-6x2+24x+9x = 15+6 33x = 21 x = 21/33 => x = 7/11
Yes.
x3 - 6x2 + 4x + 15 = (x - 3)(x - (3/2 + √29/2))(x - (3/2 - √29/2)) ⇒ roots are x = 3, x = 3/2 + √29/2 (≈ 4.19), or x = 3/2 - √29/2 (≈ -1.19)
x3 6x2-x-30
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In the absence of a proper statement of the equation I can only guess at answers. How about (2x + 3)(3x - 5) which is the factorisation of 6x2 - x - 15 = 0? In which case x = -2/3 or 5/3.
6x2 - x - 15 = 0 since (6)(-15) = (-10)(9) and -10 + 9 = -1, then factor as: [(6x + 9)/3][6x - 10)/2] = 0 (2x + 3)(3x - 5) = 0 let each factor be zero: 2x + 3 = 0 or 3x - 5 = 0 solve for x: The solutions are: x = -3/2 and x = 5/3
10x2 - 56 = 88 - 6x2 : 10x2 + 6x2 = 88 + 56 : 16x2 = 144 : x2 = 9 : x = ± 3
6x2-24=0 6(x2-4)=0 x= {-2,2}
Yes.(2x)2 = 2*2*x*xor2x2 = 2 * x * x
If after finding all the factors of the product of the highest term's coefficient and the linear term, none of the can add or subtract to give the middle term then it is not factorable. x^2 +5x+6 is factorable. 1x6; 2x3 2+3=5 so it can be factored. x^2+7x+6 is factorable but x^2+8x+6 is NOT factorable because 1 and 6 as well as 2 and 3 can't add up to the middle term.
If: 6x2+11x-10 = 0 Then: x = -5/2 and x = 2/3
x3 + 6x2 - 4x - 24 = (x + 6)(x2 - 4) = (x + 6)(x + 2)(x - 2)